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d. (4 pts) Determine the pH of a buffer containing 0.10 M HF and 0.20 M F ions (2 pts) The buffer in (d) will resist a change
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Answer #1

d) Ka for HF = 6.6 * 10^(-4)

pH = pKa + log(F-/HF) = (4 - log(6.6)) + log(2) = 3.48

e) Since it has more number of F- ions, hence it will have more buffer capacity for addition of acids, which will react with the present F- ions in solution.

f) H2SO4 is a strong acid. Hence ideally it will give 2 moles of H+ per mole of H2SO4. But the leaving of second H+ is not that fast. Hence the [H+] will be more than 1 but less than 2 times. Hence the correct answer will be "Between 0.2 and 0.4M"

g)Since Ka for HNO2 is higher, hence it will be stronger acid, so it will lead to have lesser pH value. Hence the pH value will be higher for HClO

h) The Ksp expressio for mercury sulfate will be

Hg_2SO_4\rightleftharpoons 2Hg^+ + SO_4^{2-}

K_{sp} = [Hg^+]^2[SO_4^{2-}]

Note - Post any doubts/queries in comments section.

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